Verified Answer :
% H = 100 - ( 52.14 + 34.73 )=13.13 %
assume 100 g of this compound
mass H = 13.13 g
moles H = 13.13 g / 1.008 g/mol=13
mass C = 52.14 g
moles C = 52.14 g/ / 12.011 g/mol=4
mass O = 34.73 g
moles O = 34.73 g/ 15.999 g/mol=2
the empirical formula is C4H13O2
Answer: The empirical formula is \(C_2H_6O\).
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C = 52.14 g
Mass of H = 13.13 g
Mass of O = 34.73 g
Step 1 : convert given masses into moles.
Moles of C =\(\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{52.14g}{12g/mole}=4.35moles\)
Moles of H =\(\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{13.13g}{1g/mole}=13.13moles\)
Moles of O =\(\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{34.73g}{16g/mole}=2.17moles\)
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =\(\frac{4.35}{2.17}=2\)
For H =\(\frac{13.13}{2.17}=6\)
For O =\(\frac{2.17}{2.17}=1\)
The ratio of C: H: O = 2: 6: 1
Hence the empirical formula is \(C_2H_6O\).